Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂
Answer:
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Answer:
Amount of gas still in cylinder = 28 pound
Explanation:
Given:
Amount of gas in cylinder = 50 pound
Amount of gas used in Ms. Jones system = 13 pound
Amount of gas used in client system = 9 pound
Find:
Amount of gas still in cylinder
Computation:
Amount of gas still in cylinder = Amount of gas in cylinder - Amount of gas used in Ms. Jones system - Amount of gas used in client system
Amount of gas still in cylinder = 50 - 13 - 9
Amount of gas still in cylinder = 28 pound
Answer:
2.77mpa
Explanation:
compressive strength = 20 MPa. We are to find the estimated flexure strength
We calculate the estimated flexural strength R as
R = 0.62√fc
Where fc is the compressive strength and it is in Mpa
When we substitute 20 for gc
Flexure strength is
0.62x√20
= 0.62x4.472
= 2.77Mpa
The estimated flexure strength is therefore 2.77Mpa